1363 最小公倍数之和

推式子

$$\begin{gathered} \sum _{i=1}^{n}lcm(i,n) \\ =n\sum _{i=1}^n\frac{i}{gcd(i,n)} \\ =n\sum _{d\mid n}\sum _{i=1}^n\frac{i}{d}(gcd(i,n)==d) \\ =n\sum _{d\mid n}\sum _{i=1}^{\frac{n}{d}}i(gcd(i,\frac{n}{d})==1) \\ =n\sum _{d\mid n}\frac{d\varphi (d)+(d==1)}{2} \end{gathered}$$

接下来就是筛出$\sqrt{n}$内的质数，再通过递归算出所有的因数，然后加上所有的因数对答案的贡献，具体细节在代码中描述。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e5 + 10, mod = 1e9 + 7;int prime[N], cnt;bool st[N];void init() {for(int i = 2; i < N; i++) {if(!st[i]) prime[cnt++] = i;for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;}}
}int fac[50], num[50], tot;ll ans;void solve(int pos, int n, int phi) {if(pos == tot + 1) {ans = (ans + 1ll * n * (phi + (n == 1)) / 2 % mod) % mod;//特判n为1的情况。return ;}solve(pos + 1, n, phi);//不选这个数的情况。n *= fac[pos], phi *= (fac[pos] - 1);//第一次选这个数要单独考虑，当两个数互质的时候phi[i * prime] = phi[i] * (prime - 1)solve(pos + 1, n, phi);for(int i = 1; i < num[pos]; i++) {n *= fac[pos], phi *= fac[pos];//这里就是不互质的情况了solve(pos + 1, n, phi);}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T = read();while(T--) {tot = 0;int n = read(), m = n;for(int i = 0; prime[i] * prime[i] <= n; i++) {if(n % prime[i] == 0) {fac[++tot] = prime[i], num[tot] = 0;while(n % prime[i] == 0) {n /= prime[i];num[tot]++;}}}if(n != 1) {fac[++tot] = n, num[tot] = 1;}ans = 0;solve(1, 1, 1);printf("%lld\n", ans * m % mod);}return 0;
}